Question:

A mass $M$ at rest is broken into two pieces having masses $m$ and $(M-m)$. The two masses are then separated by a distance $r$. The gravitational force between them will be the maximum when he ratio of the masses $[m:(M-m)]$ of the two parts is

Updated On: Jan 12, 2024
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The Correct Option is A

Solution and Explanation

The gravitation force between two masses,
$F=\frac{G m_{1} m_{2}}{r}$
here, $m_{1}=m$
and $m_{2} =(M-m) $
$\therefore$ F $=\frac{G m(M-m)}{r^{2}}$
For maximum gravitational force, $\frac{d F}{d m}=0$
$\therefore \frac{d}{d m}[m(M-m)]=0$
By solving, we get $m=\frac{M}{2}$
So, $\frac{m}{(M-m)}=\frac{1}{1}$
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Concepts Used:

Gravitation

In mechanics, the universal force of attraction acting between all matter is known as Gravity, also called gravitation, . It is the weakest known force in nature.

Newton’s Law of Gravitation

According to Newton’s law of gravitation, “Every particle in the universe attracts every other particle with a force whose magnitude is,

  • F ∝ (M1M2) . . . . (1)
  • (F ∝ 1/r2) . . . . (2)

On combining equations (1) and (2) we get,

F ∝ M1M2/r2

F = G × [M1M2]/r2 . . . . (7)

Or, f(r) = GM1M2/r2

The dimension formula of G is [M-1L3T-2].